Length of secant line in respect to circumference











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![circle with secant line]https://images.google.com/imgres?imgurl=http%3A%2F%2Fimage.mathcaptain.com%2Fcms%2Fimages%2F113%2Fsecant.png&imgrefurl=http%3A%2F%2Fwww.mathcaptain.com%2Fgeometry%2Fsecant.html&tbnid=mvxOwDyPBQ3svM&vet=1&docid=YUhMzW49fXk_wM&w=238&h=208 If you have a circle and you know the circumference is 30, the length of arc AB is 7, how would you find the length of the decant line AB?










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    ![circle with secant line]https://images.google.com/imgres?imgurl=http%3A%2F%2Fimage.mathcaptain.com%2Fcms%2Fimages%2F113%2Fsecant.png&imgrefurl=http%3A%2F%2Fwww.mathcaptain.com%2Fgeometry%2Fsecant.html&tbnid=mvxOwDyPBQ3svM&vet=1&docid=YUhMzW49fXk_wM&w=238&h=208 If you have a circle and you know the circumference is 30, the length of arc AB is 7, how would you find the length of the decant line AB?










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      ![circle with secant line]https://images.google.com/imgres?imgurl=http%3A%2F%2Fimage.mathcaptain.com%2Fcms%2Fimages%2F113%2Fsecant.png&imgrefurl=http%3A%2F%2Fwww.mathcaptain.com%2Fgeometry%2Fsecant.html&tbnid=mvxOwDyPBQ3svM&vet=1&docid=YUhMzW49fXk_wM&w=238&h=208 If you have a circle and you know the circumference is 30, the length of arc AB is 7, how would you find the length of the decant line AB?










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      ![circle with secant line]https://images.google.com/imgres?imgurl=http%3A%2F%2Fimage.mathcaptain.com%2Fcms%2Fimages%2F113%2Fsecant.png&imgrefurl=http%3A%2F%2Fwww.mathcaptain.com%2Fgeometry%2Fsecant.html&tbnid=mvxOwDyPBQ3svM&vet=1&docid=YUhMzW49fXk_wM&w=238&h=208 If you have a circle and you know the circumference is 30, the length of arc AB is 7, how would you find the length of the decant line AB?







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      asked 10 hours ago









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          First, find the radius of the circle: $2pi R=30$, $R=frac{15}{pi}$. Let $O$ be the center of the circle. Then $mangle{AOB}=frac{7}{R}=frac{7pi}{15}$. Observe that $triangle{AOB}$ is isosceles with $AO=OB=R$. Then $AB=2Rsin frac{7pi}{30}=frac{30}{pi}sin frac{7pi}{30}approx 6.4$






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            First, find the radius of the circle: $2pi R=30$, $R=frac{15}{pi}$. Let $O$ be the center of the circle. Then $mangle{AOB}=frac{7}{R}=frac{7pi}{15}$. Observe that $triangle{AOB}$ is isosceles with $AO=OB=R$. Then $AB=2Rsin frac{7pi}{30}=frac{30}{pi}sin frac{7pi}{30}approx 6.4$






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              First, find the radius of the circle: $2pi R=30$, $R=frac{15}{pi}$. Let $O$ be the center of the circle. Then $mangle{AOB}=frac{7}{R}=frac{7pi}{15}$. Observe that $triangle{AOB}$ is isosceles with $AO=OB=R$. Then $AB=2Rsin frac{7pi}{30}=frac{30}{pi}sin frac{7pi}{30}approx 6.4$






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                First, find the radius of the circle: $2pi R=30$, $R=frac{15}{pi}$. Let $O$ be the center of the circle. Then $mangle{AOB}=frac{7}{R}=frac{7pi}{15}$. Observe that $triangle{AOB}$ is isosceles with $AO=OB=R$. Then $AB=2Rsin frac{7pi}{30}=frac{30}{pi}sin frac{7pi}{30}approx 6.4$






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                First, find the radius of the circle: $2pi R=30$, $R=frac{15}{pi}$. Let $O$ be the center of the circle. Then $mangle{AOB}=frac{7}{R}=frac{7pi}{15}$. Observe that $triangle{AOB}$ is isosceles with $AO=OB=R$. Then $AB=2Rsin frac{7pi}{30}=frac{30}{pi}sin frac{7pi}{30}approx 6.4$







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                answered 9 hours ago









                Vasya

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